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Discrete Mathematical Structures
CS 2233 Lecture Sixteen
Prof. William Winsborough
March 19, 2009
19 March 2009 Winsborough CS 2233 Lecture 16 2
Business
•Homework 7
– 4.1: 4, 6
– 4.2: 4
• Read sections 4.1, 4.2 and 4.3
• Midterm II is in two weeks on 4/2
19 March 2009 Winsborough CS 2233 Lecture 16 3
Mathematical Induction
• How can we show that a proposition P(n) holds
for all natural numbers n ∈N?
• Proof technique called
mathematical induction:
– Basis: show P(0)
– Inductive Step: show that for all k ∈ N,
P(k) → P(k+1)
• The proposition P(k) is called the
induction
hypothesis
• (P(0) ∧ ∀k ∈ N.(P(k) → P(k+1))) → ∀n ∈ N.P(n)
19 March 2009 Winsborough CS 2233 Lecture 16 4
Example of Induction
• Theorem: P(n) ≡Σ2i= 2n+1 –1
• Proof by induction
– Basis: P( 0) ≡Σ2i= 20+1 –1 ≡2
0= 2-1, which clearly
holds
– Step: We assume P(k) ≡ Σ2i= 2k+1 –1
and show P(k+1) ≡ Σ2i= 2k+2 – 1 as follows:
Σ2i= Σ2i+ 2k+1
= (2k+1 –1) + 2
k+1 by the induction hypothesis
= 2·2k+1 –1
= 2k+2 –1
0 ≤i ≤n
0 ≤i ≤0
0 ≤i ≤k
0 ≤i ≤k+1
0 ≤i ≤k+1 0 ≤i ≤k
19 March 2009 Winsborough CS 2233 Lecture 16 5
Validity of Induction
• Induction is valid because the natural numbers
are well founded
– Definit ion: A set is well ordered if each of its subsets
has a least element
• Once basis and step are shown, the assumption
that the property fails for some values yields a
contradiction
– Assume for contradiction that
P(0) ∧ ∀k(P(k) → P(k+1)) ∧ ¬∀nP(n)
– Consider the least m ∈ N such that ¬P(m)
• Case 1: if m = 0, the contradiction is immediate
• Case 2: if m = k+1 for some k ∈ N, then by minimality of m
P(k) holds. But this, together with the step, implies that
P(k+1) ≡ P(m) holds, giving us the desired contradiction
19 March 2009 Winsborough CS 2233 Lecture 16 6
Strong Induction
• To show ∀n∈N (P(n)), the following is
sufficient:
– Basis: show P(0)
– Inductive Step: show that for all k ∈ N,
[P(0) ∧…∧ P(k)] → P(k+1)
• This gives us a stronger induction
hypothesis to use in the step
• It is valid for similar reasons to those
shown on the previous slide
